A problem from nuclear engineering: the radioactive element EP^432 decays according to the rule e^{-at} for a certain constant a. Starting from zero at time 0, it is added at a variable rate f(t). How much is in the vessel at time t? Call it x(t). It satsifies the first order linear equation
x' + ax = f(t),
and we have our methods. But think about if afresh, like this. At time t, material added at all times between 0 and t contribute to the total. We need a new variable, ranging from 0 to t: call it tau. Then (f(tau) Delta tau) EP is added between time tau and time tau + Delta tau, but it decays by the time we get to t, and contributes only
f(tau) Delta tau e^{-a(t-tau)}
then. Taking the sum and letting Delta tau tend to 0, we find
x(t) = integral_0^t f(tau) e^{a(t-tau)} d tau.
This is a uniform and explicit solution to the IVP, for any forcing term f. It combines f(t) with the decay rate e^{-at}. This way of combining two functions generalizes in a useful way to the "convolution" of two functions f and g:
(f*g)(t) = integral_0^t f(tau) g(t-tau) d tau.
Example: (cos*sin)(t) = integral_0^t cos(tau) sin(t-tau) d tau.
To compute this use the trig identities
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)
sin(A+B) + sin(A-B) = 2 sin(A)cos(B)
with A = t-tau, B = tau, so A+B = t, A-B = t - 2 tau:
(cos*sin)(t) = (1/2) integral_0^t (sin(t) + sin(t - 2 tau)) d tau.
The first term is constant and the second is integrated over a symmetric interval, so
(cos*sin)(t) = (1/2) t sin(t).
Convolution interacts with Laplace transform in a beautiful way:
L(f*g) = (Lf)(Lg) (*)
convolution in the t-domain corresponds to multiplication in the s-domain.
This is proven in the book; you reverse the order of integration in a double integral. For example,
L((1/2) t sin(t);s) = L((cos*sin)(t);s) = L(cos(t);s)L(sin(t);s)
= (s/(s^2+1))(1/(s^2+1)) = s/(s^2+1)^2.
This gives a computation of the inverse Laplace transform of s/(s^2+1)^2 different from what was assigned for homework. It illustrates the method: if the s-function is a product, you can express its inverse transform as a convolution.
The expression (*) lets us make some general conclusions about convolution. For example, it's commutative, since multiplication is. This can be checked directly from the definition as well, pretty easily. Somewhat harder is checking that it's associative as well. Does it have a unit? That is, is there a function delta such that delta*f = f for any f? If so, then L(delta) L(f) = L(f), so we must have L(delta) = 1. But we know that L(f(t);s) tends to 0 as s tends to infinity for any function f(t): so no such function can exist. Nevertheless we will find it very useful to talk about an object delta like this, and I will do this on Monday.
Can we generalize this expression of the solution of an ODE to the general linear constant coefficient case with rest initial conditions?---
p(D)x = f(t), x(0) = x'(0) = 0 (in the second order case).
Apply L : p(s)X(s) = F(s) , or X(s) = W(s)F(s) where W(s) = 1/p(s) is the "transfer function." Till now this was the stopping point, but now we can express the inverse Laplace transform as a convolution:
x = w*f, where w(t) is the function such that L(w) = W.
w is the "weight function" of the system.
For example, return to the original case x' + ax = f(t). Apply L: X(s) = W(s)F(s) where W(s) = 1/(s+a). This is the Laplace transform of w(t) = e^{-at}, so we find once again that
x(t) = e^{-at}*f(t).
For example suppose that f(t) = t. Then
x(t) = e^{-at}*t = integral_0^t (t-tau) e^{-a tau} d tau
u = t - tau, dv = e^{-a tau} ;
du = - d tau, v = e^{-a tau}/(-a);
x(t) = (t - tau) e^{-a tau}/(-a) |_0^t - integral_0^t e^{-a tau)/(-a) (-d tau)
= (t/a) + (1-e^{-at})/a^2.
Whenever you finish a piece of work like this, sit back and think about what it means. With this linear ramping up of the rate of loading, the amount of EP in the reactor also increases essentially linearly, or more precisely affinely: when t is large we get approximately (t/a) + (1/a^2).